Problem 21
Question
In Problems 9-28, find the convergence set for the given power series. Hint: First find a formula for the nth term; then use the Absolute Ratio Test. $$ 1+2 x+\frac{2^{2} x^{2}}{2 !}+\frac{2^{3} x^{3}}{3 !}+\frac{2^{4} x^{4}}{4 !}+\cdots $$
Step-by-Step Solution
Verified Answer
The power series converges for all real numbers; it has an infinite radius of convergence.
1Step 1: Identifying the nth Term
The given series is a power series, and we need to find an explicit formula for the nth term. Observing the terms, we see a pattern where each term involves \(2^n x^n / n!\). Therefore, the nth term \(a_n\) can be expressed as \(a_n = \frac{2^n x^n}{n!}\).
2Step 2: Applying the Absolute Ratio Test
The Absolute Ratio Test involves checking the limit: \(\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\). Substituting the nth term formula, we get \[ \frac{a_{n+1}}{a_n} = \frac{2^{n+1} x^{n+1} / (n+1)!}{2^n x^n / n!} = \frac{2 x}{n+1}. \] We need to evaluate \( \lim_{n \to \infty} \left| \frac{2x}{n+1} \right| \).
3Step 3: Calculating the Limit of the Ratio
Simplify the expression \( \left| \frac{2x}{n+1} \right| \). As \(n\) approaches infinity, \(\frac{1}{n+1}\) approaches zero, implying that \(\lim_{n \to \infty} \left| \frac{2x}{n+1} \right| = 0\).
4Step 4: Conclusion on Convergence
Since the limit approached zero for any real number (ignoring the value of \(x\))—this means the series converges for all \(x\). The radius of convergence is therefore infinite.
Key Concepts
Absolute Ratio Testradius of convergencenth term formulation
Absolute Ratio Test
The Absolute Ratio Test is a useful tool for determining the convergence of a power series. It involves checking the limit of the ratio of consecutive terms. For a power series with a general term denoted as \(a_n\), we consider the expression
\[\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\]If this limit is less than 1, the series converges absolutely. If the limit is greater than 1, the series diverges. The test is inconclusive if the limit equals 1.
\[\frac{2 x}{n+1}\]as the ratio for successive terms. As \(n\) approaches infinity, this ratio approaches zero, indicating convergence for any real \(x\). Thus, the power series converges absolutely for all values of \(x\).
\[\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|\]If this limit is less than 1, the series converges absolutely. If the limit is greater than 1, the series diverges. The test is inconclusive if the limit equals 1.
- This test helps determine where the series converges based on the variable \(x\).
- It gives insight into the behavior of the series as \(n\) grows larger.
\[\frac{2 x}{n+1}\]as the ratio for successive terms. As \(n\) approaches infinity, this ratio approaches zero, indicating convergence for any real \(x\). Thus, the power series converges absolutely for all values of \(x\).
radius of convergence
The radius of convergence informs us about the set of \(x\) values for which a power series converges. In the context of the Absolute Ratio Test used earlier, the radius can be identified by seeking the range of values that result in convergence.
- When the limit of the ratio is zero, the radius of convergence is said to be infinite.
- In this case, it means the series converges for every real number \(x\).
nth term formulation
Determining the formula for the \(n\)th term is crucial in analyzing a power series. The \(n\)th term formulation allows us to utilize various convergence tests. In the given series, we observe the pattern
\[1, 2x, \frac{2^2x^2}{2!}, \frac{2^3x^3}{3!}, \ldots\]From this, the general term is identified as
\[a_n = \frac{2^n x^n}{n!}\]This representation is vital for applying the Ratio Test, as it provides the iterative elements necessary to explore the limits of the converging sequence.
\[1, 2x, \frac{2^2x^2}{2!}, \frac{2^3x^3}{3!}, \ldots\]From this, the general term is identified as
\[a_n = \frac{2^n x^n}{n!}\]This representation is vital for applying the Ratio Test, as it provides the iterative elements necessary to explore the limits of the converging sequence.
- The \(n\)th term lets us establish the parameter that varies with \(x\).
- It forms the basis for mathematical manipulations and further calculations.
Other exercises in this chapter
Problem 20
In Problems 13–30, classify each series as absolutely convergent, conditionally convergent, or divergent. $$ \sum_{n=2}^{\infty}(-1)^{n} \frac{1}{\sqrt{n^{2}-1}
View solution Problem 20
In Problems 15-20, write the given decimal as an infinite series, then find the sum of the series, and finally, use the result to write the decimal as a ratio o
View solution Problem 21
In Problems 21-30, find an explicit formula \(a_{n}=\) for each sequence, determine whether the sequence converges or diverges, and, if it converges, find \(\li
View solution Problem 21
In Problems 19-24, find the Taylor series in \(x-\) a through the term \((x-a)^{3}\). $$ \cos x, a=\frac{\pi}{3} $$
View solution